DEV Community

Shashank Shekhar
Shashank Shekhar

Posted on

Rings and Rods | Leetcode Problem Solution Using Bit Manipulation

Problem Statement

There are n rings and each ring is either red, green, or blue. The rings are distributed across ten rods labeled from 0 to 9.

You are given a string rings of length 2n that describes the n rings that are placed onto the rods. Every two characters in rings forms a color-position pair that is used to describe each ring where:

  • The first character of the ith pair denotes the ith ring's color ('R', 'G', 'B').
  • The second character of the ith pair denotes the rod that the ith ring is placed on ('0' to '9').

For example, "R3G2B1" describes n == 3 rings: a red ring placed onto the rod labeled 3, a green ring placed onto the rod labeled 2, and a blue ring placed onto the rod labeled 1.

Return the number of rods that have all three colors of rings on them.

Sample Input
B0B6G0R6R0R6G9

Sample Output
1

Explanation

  • The rod labeled 0 holds 3 rings with all colors: red, green, and blue.
  • The rod labeled 6 holds 3 rings, but it only has red and blue.
  • The rod labeled 9 holds only a green ring. Thus, the number of rods with all three colors is 1.

Link to question: https://leetcode.com/problems/rings-and-rods/

Solution (C++)

class Solution {
    public int countPoints(String rings) {
        int count, n=rings.length();
        int rods[]=new int[10];
        Arrays.fill(rods, 0);
        for(int i=0; i<n; i+=2) {
            char color=rings.charAt(i);
            int rod=rings.charAt(i+1)-48;
            switch(color) {
                case 'R':
                    rods[rod]=rods[rod] | 1;
                    break;
                case 'G':
                    rods[rod]=rods[rod] | 2;
                    break;
                case 'B':
                    rods[rod]=rods[rod] | 4;
                    break;
            }
        }
        count=0;
        for(int item: rods)
            count+=item==7?1:0;
        return count;
    }
};
Enter fullscreen mode Exit fullscreen mode

Top comments (0)